+0  
 
0
945
10
avatar+262 

=0

 Jul 13, 2015

Best Answer 

 #4
avatar+26364 
+10

$$\small{\text{$
\begin{array}{rcl}
\boxed{~~ax^2+bx+c=0 \qquad
x_{1,2} = \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} ~~}
\end{array}
$}}$$

 

$$\small{\text{$
\begin{array}{rcl}
a=m ~~b=-1~~c=-\dfrac{m-2}{4}\\\\
x_{1,2} &=& \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1^2-4m\left(-\dfrac{m-2}{4}\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m\left(m-2\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m^2-2m} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1-2m+m^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{(1-m)^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 1 + (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 2-m }{2m} \\\\
x_1 &=& \dfrac{ 2 }{2m} - \dfrac{ m }{2m} \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{ \dfrac{ 1 }{m} - \dfrac{ 1 }{2} } \\\\\\
x_2 &=& \dfrac{ 1 - (1-m) }{2m} \\\\
x_2 &=& \dfrac{ 1 - 1 + m)}{2m} \\\\
x_2 &=& \dfrac{ m }{2m} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{\dfrac{ 1 }{2}}
\end{array}
$}}$$

 

.
 Jul 13, 2015
 #1
avatar+14538 
+5

$${{\mathtt{mx}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\frac{\left({\mathtt{m}}{\mathtt{\,-\,}}{\mathtt{2}}\right)}{{\mathtt{4}}}} = {\mathtt{0}}$$

 

integer solutions :   m = -2       x = 1

                               m = 2        x = 0

 Jul 13, 2015
 #2
avatar+33603 
+10

geno has already answered this here: http://web2.0calc.com/questions/help-me-please_38 

.

 Jul 13, 2015
 #3
avatar+118587 
+5

Sabi92, if you do not understand Geno's answer, you need to post questions about it on the original thread. :)

 Jul 13, 2015
 #4
avatar+26364 
+10
Best Answer

$$\small{\text{$
\begin{array}{rcl}
\boxed{~~ax^2+bx+c=0 \qquad
x_{1,2} = \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} ~~}
\end{array}
$}}$$

 

$$\small{\text{$
\begin{array}{rcl}
a=m ~~b=-1~~c=-\dfrac{m-2}{4}\\\\
x_{1,2} &=& \dfrac{ -b\pm \sqrt{b^2-4ac} }{2a} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1^2-4m\left(-\dfrac{m-2}{4}\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m\left(m-2\right) } }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1+m^2-2m} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{1-2m+m^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm \sqrt{(1-m)^2} }{2m} \\\\
x_{1,2} &=& \dfrac{ 1\pm (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 1 + (1-m) }{2m} \\\\
x_1 &=& \dfrac{ 2-m }{2m} \\\\
x_1 &=& \dfrac{ 2 }{2m} - \dfrac{ m }{2m} \\\\
\mathbf{x_1} & \mathbf{=} & \mathbf{ \dfrac{ 1 }{m} - \dfrac{ 1 }{2} } \\\\\\
x_2 &=& \dfrac{ 1 - (1-m) }{2m} \\\\
x_2 &=& \dfrac{ 1 - 1 + m)}{2m} \\\\
x_2 &=& \dfrac{ m }{2m} \\\\
\mathbf{x_2} & \mathbf{=} & \mathbf{\dfrac{ 1 }{2}}
\end{array}
$}}$$

 

heureka Jul 13, 2015
 #5
avatar+14538 
+5

 x = $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$      und      x = $${\frac{{\mathtt{1}}}{{\mathtt{m}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

 

Gruß radix !

 

 Jul 13, 2015
 #6
avatar+262 
0

Oh sorry i haven't seen his answer

 Jul 13, 2015
 #7
avatar+118587 
+5

Why didn't you see his answer?

The fact that you had an answer would have been clearly displayed in your watch list.  :/

 Jul 13, 2015
 #8
avatar+262 
0

yes but i think i confused it with other answer and reloaded the browser whithout noticed that its exactly the answer i need

 Jul 13, 2015
 #9
avatar+262 
+5

 i think its because my watchlist is a bit messy with all of my posts and i should remove some of them from my watchlist

 Jul 13, 2015
 #10
avatar+118587 
+5

Ok, it is not a problem sabi,  

I am sure you will get yourself better organised.   

 Jul 14, 2015

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